CHAPTER 8 – QUADRILATERALS (Class 9, NCERT/CBSE)
Part A — 1-Mark Questions (20) with Solutions
Q1. Define a parallelogram.
A quadrilateral in which both pairs of opposite sides are parallel.
Q2. State Theorem 8.1.
A diagonal of a parallelogram divides it into two congruent triangles.
Q3. In a parallelogram, opposite sides are…?
Equal and parallel: \(AB=CD\) and \(BC=AD\).
Q4. In a parallelogram, opposite angles are…?
Equal: \(\angle A=\angle C,\ \angle B=\angle D\).
Q5. Diagonals of a parallelogram…?
Bisect each other.
Q6. If diagonals of a quadrilateral bisect each other, it is a…?
Parallelogram (converse of Theorem 8.6).
Q7. Diagonals of a rectangle are…?
Equal and bisect each other.
Q8. Diagonals of a rhombus are…?
Perpendicular and bisect each other.
Q9. Diagonals of a square are…?
Equal, perpendicular, and bisect each other.
Q10. A line segment joining mid-points of two sides of a triangle is…?
Parallel to the third side and equals half of it: \(EF\parallel BC,\ EF=\tfrac12 BC\).
Q11. Name the theorem that is converse to Theorem 8.8.
The Mid-point Theorem converse (Theorem 8.9).
Q12. In a parallelogram \(ABCD\), if \(AC\) is a diagonal, which triangles are congruent?
\(\triangle ABC \cong \triangle CDA\) (ASA).
Q13. If each pair of opposite sides of a quadrilateral is equal, then it is a…?
Parallelogram (Theorem 8.3).
Q14. In a trapezium, how many pairs of parallel sides exist?
Exactly one pair.
Q15. A rectangle is a parallelogram with…?
One (hence each) angle \(=90^\circ\).
Q16. A rhombus is a parallelogram with…?
All sides equal.
Q17. The sum of interior angles of any quadrilateral is…?
\(360^\circ\).
Q18. In triangle \(ABC\), if \(E,F\) are mid-points of \(AB,AC\), then \(EF=\ ?\)
\(EF=\dfrac{1}{2}BC\).
Q19. The point where diagonals of a parallelogram meet is the…?
Mid-point of each diagonal (common intersection bisects both).
Q20. If diagonals of a parallelogram are equal, the parallelogram is a…?
Rectangle.
Part B — 2-Mark Questions (20) with Solutions
Q1. Prove Theorem 8.1 quickly.
In parallelogram \(ABCD\), diagonal \(AC\) gives \(\angle BCA=\angle DAC\) and \(\angle BAC=\angle DCA\) (alternate interior). With common \(AC\), by ASA: \(\triangle ABC\cong\triangle CDA\).
Q2. Show that opposite sides of a parallelogram are equal.
From Q1, CPCT on congruent triangles yields \(AB=CD\) and \(BC=AD\).
Q3. If in quadrilateral \(ABCD\), \(AB=CD\) and \(BC=AD\), prove it is a parallelogram.
Diagonal \(AC\): \(\triangle ABC\cong\triangle CDA\) by SSS. Then corresponding angles are equal \(\Rightarrow\) alternate interior angles equal \(\Rightarrow AB\parallel CD\) and \(BC\parallel AD\). Hence parallelogram.
Q4. Prove that opposite angles of a parallelogram are equal.
From \(\triangle ABC\cong\triangle CDA\), CPCT gives \(\angle ABC=\angle CDA\) and \(\angle BAC=\angle DCA\). Thus opposite angles equal.
Q5. Prove the converse: if opposite angles of a quadrilateral are equal, it is a parallelogram.
Use angle sum of quadrilateral and properties of parallel lines: equal opposite angles imply each pair is supplementary with adjacent angles, forcing opposite sides parallel.
Q6. Show that diagonals of a parallelogram bisect each other.
In \(ABCD\) with diagonals intersecting at \(O\), from \(\triangle AOB\cong\triangle COD\) (ASA), \(AO=OC\) and \(BO=OD\). Thus they bisect.
Q7. If diagonals of a quadrilateral bisect each other, show it’s a parallelogram.
Given \(AO=OC\) and \(BO=OD\), get \(\triangle AOB\cong\triangle COD\) (SSS). Hence \(\angle ABO=\angle CDO\Rightarrow AB\parallel CD\). Similarly \(BC\parallel AD\). Hence a parallelogram.
Q8. Prove each angle of a rectangle is \(90^\circ\).
Rectangle is a parallelogram with one right angle; adjacent interior angles on a transversal are supplementary, giving all \(90^\circ\).
Q9. Show that diagonals of a rhombus are perpendicular.
In rhombus \(ABCD\), \(AB=BC=CD=DA\). In \(\triangle AOD\) and \(\triangle COD\): \(OA=OC\) (bisected), \(OD\) common, \(AD=CD\). SSS \(\Rightarrow\) congruent \(\Rightarrow \angle AOD=\angle COD\). Linear pair sums \(180^\circ\Rightarrow \angle AOD=90^\circ\).
Q10. Two parallel lines cut by a transversal: show that angle-bisector quadrilateral is a rectangle.
Using alternate interior angles and halving, opposite sides are parallel; one interior angle becomes \(90^\circ\). Hence rectangle (as in Example 4).
Q11. Show that angle bisectors of a parallelogram form a rectangle.
Sum of two adjacent halved angles in triangle equals \(90^\circ\), repeating cyclically gives all four right angles; hence rectangle (Example 5).
Q12. State and prove Mid-point Theorem.
If \(E\) and \(F\) are mid-points of \(AB,AC\) in \(\triangle ABC\), then \(EF\parallel BC\) and \(EF=\tfrac12 BC\). Use parallelogram construction via a congruent triangle as in text.
Q13. State and prove converse of Mid-point Theorem.
Line through mid-point \(E\) of one side parallel to another side meets third side at its mid-point. Use ASA congruence of suitable triangles to conclude halving.
Q14. In \(\triangle ABC\), \(D,E,F\) are mid-points of \(AB,BC,CA\). Show the four small triangles are congruent.
Using Mid-point Theorem thrice, get three parallelograms inside and diagonals bisect; all four sub-triangles are congruent (Example 6).
Q15. Three parallel lines \(l,m,n\) cut equal intercepts on one transversal; prove they cut equal intercepts on any other transversal.
Join endpoints, use converse mid-point argument twice (Example 7) to show second transversal’s intercepts are equal.
Q16. If diagonals of a parallelogram are equal, show it is a rectangle.
In \(ABCD\), with \(AC=BD\) and diagonals bisecting, triangles adjacent to a side become congruent giving a right angle; hence rectangle (Exercise 8.1 Q1 idea).
Q17. In parallelogram \(ABCD\), show that the mid-point of any diagonal is the mid-point of the other.
Diagonals bisect each other at same point \(O\); \(O\) is mid-point of both \(AC\) and \(BD\).
Q18. In a trapezium \(AB\parallel CD\), show that angle \(A\) and angle \(D\) are supplementary.
With \(AD\) as transversal, interior angles on same side are supplementary: \(\angle A+\angle D=180^\circ\).
Q19. Prove that the diagonals of a square have equal length and are perpendicular.
Square is both rectangle (diagonals equal) and rhombus (diagonals perpendicular). Hence both properties hold.
Q20. In \(\triangle ABC\), mid-point \(M\) of hypotenuse \(AB\) (right-angled at \(C\)). Show \(CM=AM=BM\).
Right triangle median to hypotenuse equals half the hypotenuse and is equidistant from all vertices: by RHS congruence of \(\triangle AMC\) and \(\triangle BMC\).
Part C — 3-Mark Questions (20) with Solutions
Q1. In parallelogram \(ABCD\), prove \(\triangle ABC\cong \triangle CDA\) and deduce \(AB=CD\) and \(BC=AD\).
Alternate interior angles about \(AC\) are equal; with common \(AC\), ASA gives congruence. CPCT \(\Rightarrow\) opposite sides equal.
Q2. Prove Theorem 8.3 (converse of opposite sides equal).
Given \(AB=CD\) and \(BC=AD\). With diagonal \(AC\), \(\triangle ABC\cong\triangle CDA\) (SSS). Hence corresponding angles equal \(\Rightarrow\) both pairs of opposite sides are parallel; thus a parallelogram.
Q3. Show that if each pair of opposite angles in a quadrilateral are equal, then diagonals bisect each other.
Opposite sides become parallel \(\Rightarrow\) parallelogram \(\Rightarrow\) diagonals bisect each other (Theorem 8.6).
Q4. In rhombus \(ABCD\), prove diagonals are perpendicular and bisect vertex angles.
Perpendicular: as in 2-mark Q9. Also, congruence of adjacent triangles shows \(\angle BAD=\angle CAD\) and similarly at other vertices; hence they bisect angles.
Q5. For a rectangle, prove diagonals are equal using congruence.
In \(ABCD\) with \(\angle A=\angle B=90^\circ\): \(\triangle ABC\) and \(\triangle BAD\) have \(AB\) common, right angles, and \(BC=AD\) (opposite sides). RHS \(\Rightarrow AC=BD\).
Q6. In square \(ABCD\), prove that the angle-bisector figure formed inside is also a square.
All angles \(90^\circ\); angle bisectors each \(45^\circ\). The formed quadrilateral has equal sides and right angles \(\Rightarrow\) square.
Q7. If a line through mid-point \(E\) of \(AB\) in \(\triangle ABC\) meets \(AC\) at \(F\) with \(EF\parallel BC\), show \(F\) is mid-point of \(AC\) and \(EF=\tfrac12 BC\).
By converse Mid-point Theorem, \(F\) is mid-point of \(AC\). Then by Mid-point Theorem, \(EF\parallel BC\) and \(EF=\tfrac12 BC\).
Q8. In parallelogram \(ABCD\), prove that triangles formed by one diagonal have equal areas.
Diagonal \(AC\) splits into triangles on the same base with same altitude, or use congruence. Hence areas equal.
Q9. In triangle \(ABC\), \(D,E\) are mid-points of \(AB,BC\). Show \(DE\parallel AC\) and find ratio of areas \([ \triangle ADE ] : [ \triangle ABC]\).
\(DE\parallel AC\) by Mid-point Theorem. Similar triangles give linear scale \(1:2\) so area ratio \(=\left(\tfrac12\right)^2=\tfrac14\).
Q10. In trapezium \(AB\parallel CD\), show that the line joining mid-points of non-parallel sides is parallel to bases and equals half the sum of bases.
Use vector/coordinate or triangle decomposition to show segment length \(=\tfrac12(AB+CD)\) and parallel to bases.
Q11. Prove: diagonals of a kite are perpendicular, if the equal sides are adjacent.
Congruence of two pairs of adjacent triangles with equal sides shows one diagonal is axis of symmetry, hence perpendicular to the other.
Q12. In a parallelogram, the angle between one side and a diagonal equals the angle between the opposite side and the other diagonal. Prove.
Use parallelism to transfer angles and CPCT on congruent halves across a diagonal.
Q13. Show that the segment joining mid-points of diagonals of a trapezium is parallel to the bases.
Use coordinates: let bases be horizontal, compute mid-points; slope turns out equal to base slope.
Q14. If one angle of a parallelogram is acute and adjacent one is obtuse, prove the larger diagonal lies opposite to the obtuse angle.
Compare triangles formed by diagonal using Law of Cosines or angle-side comparisons; the diagonal opposite obtuse angle is longer.
Q15. Prove that the medians of a triangle form a triangle (the medial triangle) similar to the original.
Midpoints connected give \(DE\parallel AC\), etc. All three sides are half of the corresponding sides \(\Rightarrow\) similar with ratio \(1:2\).
Q16. In parallelogram \(ABCD\), \(E\) on \(AB\) and \(F\) on \(CD\) with \(AE=CF\). Show \(EF\parallel AD\).
Translate equal segments along parallels; corresponding triangles are congruent or use vectors to show direction equals \(AD\).
Q17. Show: the perpendicular bisectors of the sides of any parallelogram are concurrent.
They meet at the circumcenter of each of the congruent triangles formed by a diagonal—same point as midpoint of diagonals’ intersection.
Q18. If a parallelogram has one right angle, prove it is a rectangle and its diagonals are equal.
Adjacent angles supplementary force all angles \(90^\circ\). Then use RHS/ASA to show \(AC=BD\).
Q19. In triangle \(ABC\), \(E,F\) are on \(AC,AB\) such that \(BE\perp AC\) and \(CF\perp AB\). If \(BE=CF\), prove \(AB=AC\).
Right triangles \(ABE\) and \(ACF\) have equal hypotenuse legs via RHS reasoning (with common \(\angle A\)), giving \(AB=AC\).
Q20. Prove that the mid-point quadrilateral of any quadrilateral is a parallelogram.
Join consecutive mid-points; opposite sides turn out parallel by Mid-point Theorem applied in the two triangles formed by a diagonal.
Exercise 8.1 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. If the diagonals of a parallelogram are equal, show that it is a rectangle. | In parallelogram \(ABCD\), suppose \(AC=BD\). Consider \(\triangle ABC\) and \(\triangle BAD\): \(AB\) common, \(BC=AD\) (opp. sides), and \(AC=BD\) (given). By SSS, triangles are congruent \(\Rightarrow \angle ABC=\angle BAD\). Adjacent angles of a parallelogram are supplementary, so each must be \(90^\circ\). Hence a rectangle. |
| Q2. Show that the diagonals of a square are equal and bisect each other at right angles. | A square is both a rectangle and a rhombus. From rectangle: diagonals equal and bisect. From rhombus: diagonals are perpendicular bisectors. Hence equal, bisect, and at \(90^\circ\). |
| Q3. In parallelogram \(ABCD\), diagonal \(AC\) bisects \(\angle A\). Show (i) it bisects \(\angle C\) as well, (ii) \(ABCD\) is a rhombus. |
(i) From \(\triangle ABC\cong \triangle CDA\) (Theorem 8.1), CPCT gives \(\angle BAC=\angle DCA\), so \(AC\) also bisects \(\angle C\). (ii) CPCT yields \(AB=CD\) and \(BC=AD\). With angle bisector along a diagonal, adjacent sides become equal: \(AB=BC\). Thus all sides equal \(\Rightarrow\) rhombus. |
| Q4. Rectangle \(ABCD\) where diagonal \(AC\) bisects \(\angle A\) and \(\angle C\). Show (i) \(ABCD\) is a square, (ii) diagonal \(BD\) bisects \(\angle B\) and \(\angle D\). |
(i) In rectangle, \(AB\parallel CD\), \(BC\parallel AD\); if \(AC\) bisects both vertex angles, congruent triangles along \(AC\) give \(AB=BC\). Rectangle with adjacent sides equal \(\Rightarrow\) square. (ii) By symmetry of square, diagonals are angle bisectors; or show \(\triangle ABD\cong \triangle CBD\) (SSS) to get \(\angle ABD=\angle DBC\). |
| Q5. In parallelogram \(ABCD\), points \(P,Q\) on \(BD\) with \(DP=BQ\). Prove: (i) \(\triangle APD\cong \triangle CQB\); (ii) \(AP=CQ\); (iii) \(\triangle AQB\cong \triangle CPD\); (iv) \(AQ=CP\); (v) \(APCQ\) is a parallelogram. |
(i) In \(\triangle APD\) & \(\triangle CQB\): \(AD=CB\) (opp. sides), \(DP=BQ\) (given), and \(\angle APD=\angle CQB\) (vertically opposite). SAS \(\Rightarrow\) congruent. (ii) CPCT: \(AP=CQ\). (iii) Similarly, using \(AQ\) vs \(CP\) and equal segments on diagonal, get SAS \(\Rightarrow\) congruent. (iv) CPCT: \(AQ=CP\). (v) Now \(AP\parallel CQ\) and \(AQ\parallel CP\) from corresponding equal angles; or opposite sides equal and parallel \(\Rightarrow APCQ\) is a parallelogram. |
| Q6. In parallelogram \(ABCD\), \(AP\) and \(CQ\) are perpendiculars from \(A\) and \(C\) to diagonal \(BD\). Show (i) \(\triangle APB\cong \triangle CQD\); (ii) \(AP=CQ\). |
(i) Right triangles at \(P,Q\) with \( \angle APB=\angle CQD=90^\circ\). Also \(\angle PBA=\angle QDC\) (alternate interior as \(AB\parallel CD\)) and \(BD\) common as hypotenuse in the right-triangle pairs. RHS \(\Rightarrow\) congruent. (ii) CPCT gives \(AP=CQ\). |
| Q7. Trapezium \(AB\parallel CD\) with \(AD=BC\). Show (i) \(\angle A=\angle B\), (ii) \(\angle C=\angle D\), (iii) \(\triangle ABC\cong \triangle BAD\), (iv) \(AC=BD\). | Extend \(AB\) and draw through \(C\) a line \(\parallel AD\) to meet \(AB\) produced at \(E\). Then alternate interior and isosceles style relations give (i) and (ii). For (iii), use SAS on \(\triangle ABC\) and \(\triangle BAD\): \(AD=BC\) (given), included angles equal from (i), and \(AB\) common. CPCT gives (iv) \(AC=BD\). |
Exercise 8.2 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. In quadrilateral \(ABCD\), \(P,Q,R,S\) are mid-points of \(AB,BC,CD,DA\). \(AC\) is a diagonal. Show: (i) \(SR\parallel AC\) and \(SR=\tfrac12 AC\); (ii) \(PQ=SR\); (iii) \(PQRS\) is a parallelogram. |
(i) In \(\triangle ADC\), \(S\) and \(R\) are mid-points of \(AD,CD\). By Mid-point Theorem, \(SR\parallel AC\) and \(SR=\tfrac12 AC\). (ii) Similarly in \(\triangle ABC\), \(P,Q\) mid-points \(\Rightarrow PQ\parallel AC\) and \(PQ=\tfrac12 AC\). Hence \(PQ=SR\). (iii) With \(PQ\parallel SR\) and also \(PS\parallel QR\) (by same reasoning in the other triangle), opposite sides are parallel \(\Rightarrow PQRS\) is a parallelogram. |
| Q2. \(ABCD\) is a rhombus and \(P,Q,R,S\) mid-points of \(AB,BC,CD,DA\). Show \(PQRS\) is a rectangle. | By Q1, \(PQRS\) is a parallelogram. In a rhombus, diagonals are perpendicular; mid-point joins are parallel to diagonals, hence adjacent sides of \(PQRS\) are perpendicular. Thus all angles \(90^\circ\Rightarrow\) rectangle. |
| Q3. \(ABCD\) is a rectangle and \(P,Q,R,S\) mid-points of \(AB,BC,CD,DA\). Show \(PQRS\) is a rhombus. | From Q1, \(PQRS\) is a parallelogram with all sides \(=\tfrac12\) of the same diagonal \(AC\). Hence all four sides equal \(\Rightarrow\) rhombus. |
| Q4. Trapezium \(AB\parallel DC\), diagonal \(BD\), \(E\) mid-point of \(AD\). Through \(E\), draw line \(\parallel AB\) meeting \(BC\) at \(F\). Show \(F\) is mid-point of \(BC\). | Consider \(\triangle ABD\): \(E\) is midpoint of \(AD\) and \(EF\parallel AB\). By converse Mid-point Theorem, \(F\) is midpoint of \(BD\). In \(\triangle BCD\), since \(F\) is midpoint of \(BD\) and \(EF\parallel AB\parallel DC\), again by Mid-point Theorem, \(F\) is midpoint of \(BC\). |
| Q5. In parallelogram \(ABCD\), \(E\) and \(F\) are mid-points of \(AB\) and \(CD\). Show that \(AF\) and \(EC\) trisect diagonal \(BD\). | Let \(AF\) meet \(BD\) at \(X\) and \(EC\) meet \(BD\) at \(Y\). Using similar triangles from intersecting transversals and mid-point properties, obtain ratios \(BX:XD=1:2\) and \(BY:YD=2:1\). Hence \(X\) and \(Y\) divide \(BD\) into three equal parts: \(BX=XY=YD\). |
| Q6. \(\triangle ABC\) right-angled at \(C\). Line through mid-point \(M\) of hypotenuse \(AB\) parallel to \(BC\) meets \(AC\) at \(D\). Show (i) \(D\) is mid-point of \(AC\); (ii) \(MD\perp AC\); (iii) \(CM=MA=\tfrac12 AB\). |
(i) In \(\triangle ACB\), \(M\) midpoint of \(AB\) and \(MD\parallel CB\). By converse Mid-point Theorem, \(D\) is midpoint of \(AC\). (ii) In a right triangle, midpoint of hypotenuse is equidistant from all vertices; \(M\) is circumcenter, so \(MC=MA=MB\). Radius to midpoint \(D\) on \(AC\) implies \(MD\) is perpendicular to chord \(AC\). (iii) Median to hypotenuse equals half the hypotenuse: \(CM=AM=BM=\dfrac{AB}{2}\). |
All set! 🎉
Questions are in red and solutions in green. Math is rendered with MathJax for perfect clarity on mobile.